Problem: A certain circle can be represented by the following equation. $x^2+y^2+6x+2y-6=0$ What is the center of this circle ? $($
The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+6x+2y-6&=0\\\\ x^2+y^2+6x+2y&=6\\\\ (x^2+6x)+(y^2+2y)&=6 \text{(rearrange terms)}\\\\ (x^2+6x{+9})+(y^2+2y{+1})&=6{+9}{+1}\end{aligned}$ Notice that we must add ${9}$ and ${1}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 9 and 1?] Writing the equation in standard form $\begin{aligned}(x^2+6x{+9})+(y^2+2y{+1})&=6{+9}{+1}\\\\ (x+3)^2+(y+1)^2&=16\\\\ (x-(-3))^2+(y-(-1))^2&=4^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(-3,-1)$ and has a radius of $4$ units. Summary The circle is centered at $(-3,-1)$. The circle has a radius of $4$ units.